Question

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes (1st and 4rd ) to the product of means (2nd
and 3rd ) is 5:6.

Answer 1

Answer:

let the four parts be a-3d, a-d,a+d, a+3d.

given, (a-3d)+(a-d)+(a+d)+(a+3d)=56

⇒    4a=56

⇒    a = 14

also, (a-3d)(a+3d)/(a-d)(a+d)=5/6

⇒   $a^{2}-9d^2/a^2-d^2=5/6$

⇒   6(196- 9d²)= 5(196-d²)                                               [∵ a = 14]

⇒   6×196 -54d² = 5×196 - 5d²

⇒   49d² = 6× 196 -5× 196

⇒   d² = 4

⇒   d = ±2

∴ required parts are 14 - 3×2, 14-2, 14 + 2, 14+3×2

or 14-3(-2), 14-(-2), 14 + (-2), 14+3(-2)

i.e. , 8, 12, 16 , 20 or 20, 16, 12, 8

Step-by-step explanation:

Answer 2

Question :--- Divide 56 in four parts in A.P. such that the ratio of the product of their extremes (1st and 4rd ) to the product of means (2nd and 3rd ) is 5:6.

Solution :---

Let the Four Parts be :-- (a-3d) , (a-d) , (a+d) and (a+3d)

it is given that , their sum is 56,

so,

→ (a-3d) + (a-d) + (a+d) + (a+3d) = 56

→ 4a = 56

Dividing both sides by 4 , we get,

→ a = 14 .

Putting value we get,

→ a1 = a -3d = 14-3d

→ a2 = a-d = 14-d

→ a3 = a+d = 14+d

→ a4 = a+3d = 14+3d

_____________________________

Now, it is given that,

( a1*a4) : (a2*a3) = 5:6

or, we can say that,

→ (a1*a4)/(a2*a3) = 5/6

______________________________

Putting values now we get,

→ (14-3d)(14+3d) / (14-d)(14+d) = 5/6

using (a-b)(a+b) = a² - b² in LHS both numerator and denominator now, we get,

→ (196 - 9d²) / (196 - d²) = 5/6

cross- Multiply now,

→ 6*196 - 54d² = 196*5 - 5d²

→ 6*196-5*196 = -5d² + 54d²

→ 196(6-5) = 49d²

→ 196 = 49d²

→ d² = 196/49 = 4

Square - root both sides ,

→ d = 2

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Putting value we get,

→ a1 = 14-3d = 14 - 3*2 = 14 - 6 = 8

→ a2 = 14-d = 14 - 2 = 12

→ a3 = 14+d = 14 + 2 = 16

→ a4 = 14+3d = 14 + 6 = 20 .

Hence, Four parts will be 8, 12 , 16 and 20 .